实验5

task 1-1

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#include <stdio.h>
#define N 5
void input(int x[], int n);
void output(int x[], int n);
void find_min_max(int x[], int n, int *pmin, int *pmax);
int main() {
    int a[N];
    int min, max;
    printf("录入%d个数据:\n", N);
    input(a, N);
    printf("数据是: \n");
    output(a, N);
    printf("数据处理...\n");
    find_min_max(a, N, &min, &max);
    printf("输出结果:\n");
    printf("min = %d, max = %d\n", min, max);
    return 0;
}
void input(int x[], int n) {
    int i;
    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}
void output(int x[], int n) {
    int i;
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}
void find_min_max(int x[], int n, int *pmin, int *pmax) {
    int i;
    *pmin = *pmax = x[0];
    for(i = 0; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}

屏幕截图 2025-12-12 131528

1 找出最大值和最小值
2 均指向数组x的首项地址

task1-2

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#include <stdio.h>
#define N 5
void input(int x[], int n);
void output(int x[], int n);
int *find_max(int x[], int n);
int main() {
    int a[N];
    int *pmax;
    printf("录入%d个数据:\n", N);
    input(a, N);
    printf("数据是: \n");
    output(a, N);
    printf("数据处理...\n");
    pmax = find_max(a, N);
    printf("输出结果:\n");
    printf("max = %d\n", *pmax);
    return 0;
}
void input(int x[], int n) {
    int i;
    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}
void output(int x[], int n) {
    int i;
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}
int *find_max(int x[], int n) {
    int max_index = 0;
    int i;
    for(i = 0; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    return &x[max_index];
}

屏幕截图 2025-12-12 132358

1 找出最大值
2 可以

task2-1

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#include <stdio.h>
#include <string.h>
#define N 80
int main() {
    char s1[N] = "Learning makes me happy";
    char s2[N] = "Learning makes me sleepy";
    char tmp[N];
    printf("sizeof(s1) vs. strlen(s1): \n");
    printf("sizeof(s1) = %d\n", sizeof(s1));
    printf("strlen(s1) = %d\n", strlen(s1));
    printf("\nbefore swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    printf("\nswapping...\n");
    strcpy(tmp, s1);
    strcpy(s1, s2);
    strcpy(s2, tmp);
    printf("\nafter swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    return 0;
}

屏幕截图 2025-12-12 132855

1 80 数组s1的空间 数组s1中的字数
2 不可以 报错了
3 交换

task2-2

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#include <stdio.h>
#include <string.h>
#define N 80
int main() {
    char *s1 = "Learning makes me happy";
    char *s2 = "Learning makes me sleepy";
    char *tmp;
    printf("sizeof(s1) vs. strlen(s1): \n");
    printf("sizeof(s1) = %d\n", sizeof(s1));
    printf("strlen(s1) = %d\n", strlen(s1));
    printf("\nbefore swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    printf("\nswapping...\n");
    tmp = s1;
    s1 = s2;
    s2 = tmp;
    printf("\nafter swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    return 0;
}

屏幕截图 2025-12-12 134135

1 字符串Learning makes me happy 不知 字符个数
2 可以 数组变成指针
3 地址 不

task3

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#include <stdio.h>
int main() {
    int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;
    int *ptr1;   // 指针变量,存放int类型数据的地址
    int(*ptr2)[4]; // 指针变量,指向包含4个int元素的一维数组
    printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
    for (i = 0; i < 2; ++i) {
        for (j = 0; j < 4; ++j)
            printf("%d ", x[i][j]);
        printf("\n");
    }
    printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
    for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
        printf("%d ", *ptr1);
        if ((i + 1) % 4 == 0)
            printf("\n");
    }
    printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
    for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr2 + j));
        printf("\n");
    }
    return 0;
}

屏幕截图 2025-12-12 135949

数组所在地址 数组名

task4

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#include <stdio.h>
#define N 80
void replace(char *str, char old_char, char new_char); // 函数声明
int main() {
    char text[N] = "Programming is difficult or not, it is a question.";
    printf("原始文本: \n");
    printf("%s\n", text);
    replace(text, 'i', '*'); // 函数调用 注意字符形参写法,单引号不能少
    printf("处理后文本: \n");
    printf("%s\n", text);
    return 0;
}
// 函数定义
void replace(char *str, char old_char, char new_char) {
    int i;
    while(*str) {
        if(*str == old_char)
            *str = new_char;
        str++;
    }
}

屏幕截图 2025-12-12 140552

1 替换i为*
2 可以

task 5

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#include <stdio.h>
#define N 80
char *str_trunc(char *str, char x);
int main() {
    char str[N];
    char ch;
    while(printf("输入字符串: "), gets(str) != NULL) {
        printf("输入一个字符: ");
        ch = getchar();
        printf("截断处理...\n");
        str_trunc(str, ch); // 函数调用
        printf("截断处理后的字符串: %s\n\n", str);
        getchar();
    }
    return 0;
}
char *str_trunc(char *str, char x) {
    char *p = str;  
    while (*p != '\0') {
        p++; 
        if (*p == x) {
            *p = '\0';  
        }
         
    }
    return str; 
}

屏幕截图 2025-12-12 142645

无法多组输入 多组输入

task6

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#include <stdio.h>
#include <string.h>
#define N 5
int check_id(char *str); // 函数声明
int main()
{
    char *pid[N] = {"31010120000721656X",
                    "3301061996X0203301",
                    "53010220051126571",
                    "510104199211197977",
                    "53010220051126133Y"};
    int i;
    for (i = 0; i < N; ++i)
        if (check_id(pid[i])) // 函数调用
            printf("%s\tTrue\n", pid[i]);
        else
            printf("%s\tFalse\n", pid[i]);
    return 0;
}
int check_id(char *str) {
    int i;
    if(strlen(str) != 18)
        return 0;
    if(!(str[17] >= '0' && str[17] <= '9' || str[17] == 'x'))
        return 0;
    for(i = 0;i < 17;++i)
        if(!(str[i] >= '0' && str[i] <= '9'))
            return 0;
    return 1;
}

屏幕截图 2025-12-12 145425

task7

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#include <stdio.h>
#define N 80
void encoder(char *str, int n); // 函数声明
void decoder(char *str, int n); // 函数声明
int main() {
    char words[N];
    int n;
    printf("输入英文文本: ");
    gets(words);
    printf("输入n: ");
    scanf("%d", &n);
    printf("编码后的英文文本: ");
    encoder(words, n);  // 函数调用
    printf("%s\n", words);
    printf("对编码后的英文文本解码: ");
    decoder(words, n); // 函数调用
    printf("%s\n", words);
    return 0;
}

void decoder(char *str, int n) {
    while(*str != '\0'){
    if(*str >= 'a' && *str <= 'z')
        *str = 'a' + ((*str - 'a') + 26 - n) % 26;
    if(*str >= 'A' && *str <= 'Z')
        *str = 'A' + ((*str - 'A') + 26 - n) % 26;
    str++;
    }
}

void encoder(char *str, int n) {
    while(*str  !=  '\0'){
    if(*str >= 'a' && *str <= 'z')
        *str = 'a' + ((*str - 'a') + n) % 26;
    if(*str >=  'A' && *str <= 'Z')
        *str = 'A' + ((*str - 'A') + n) % 26;
    str++;
    }
}

屏幕截图 2025-12-12 150752

task 8

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#include <stdio.h>
#include <string.h>
void sort(int n, char *s[]);
int main(int argc, char *argv[]) {
    int i;
    for(i = 1; i < argc; ++i)
        printf("hello, %s\n", argv[i]);
    return 0;
}
void sort(int n, char *s[]){
    int i, j;
    char *tmp;
    for(i = 0;i < n - 1;++i)
        for(j=0;i < n-1-i;j++)
            if(strcmp(s[j], s[j+1]) > 0 ){
                tmp = s[j];
                s[j] = s[j+1];
                s[j+1] = tmp;
            }
}

最后一步忘了

posted @ 2025-12-12 15:28  陈禹江  阅读(1)  评论(0)    收藏  举报